Question

Find the perimeter of the triangle whose vertices are the following, specified points in the plane.

Answer 1

Given:

The vertices of the triangle are (-10,-3), (1,4) and (-1,7).

To find:

The perimeter of the triangle.

Solution:

Distance formula:

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Let the vertices of the triangle are A(-10,-3), B(1,4) and C(-1,7).

Using distance formula, we get

$AB=\sqrt{(1-(-10))^2+(4-(-3))^2}$

$AB=\sqrt{(1+10)^2+(4+3)^2}$

$AB=\sqrt{(11)^2+(7)^2}$

$AB=\sqrt{121+49}$

$AB=\sqrt{170}$

Similarly,

$BC=\sqrt{\left(-1-1\right)^2+\left(7-4\right)^2}=\sqrt{13}$

$AC=\sqrt{\left(-1-\left(-10\right)\right)^2+\left(7-\left(-3\right)\right)^2} =\sqrt{181}$

Now, the perimeter of the triangle is

$Perimeter=AB+BC+AC$

$Perimeter=\sqrt{170}+\sqrt{13}+\sqrt{181}$

$Perimeter\approx 13.038+3.606+13.454$

$Perimeter=30.098$

Therefore, the perimeter of the triangle is 30.098 units.